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I have a matrix whose diagonal entries are positive whereas non-diagonal entries are negative.This matrix is also Strictly diagonally dominant.

Can we say that all elements of the inverse of this matrix is strictly positive i.e $a_{ij}$>0 .

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Yes. Scale $A$ by a positive factor and we may assume that $\max_ia_{ii}<1$. Then $B:=I-A$ is positive and $$ \sum_j|b_{ij}| =|b_{ii}|+\sum_{j\ne i}|b_{ij}| =1-a_{ii}+\sum_{j\ne i}|a_{ij}| <1 $$ for each $i$. Hence $\|B\|_\infty<1$ and we may expand $A^{-1}=(I-B)^{-1}$ as an infinite sum $I+B+B^2+\ldots$. However, as $B$ is positive, the infinite sum is positive too.

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